details

property	value
status	complete
benchmark	2.19.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.159151077271 seconds
cpu usage	0.152010995
max memory	5795840.0

stage attributes

key	value
output-size	8388
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. !plus : [o * o] --> o 0 : [] --> o double : [o] --> o s : [o] --> o sqr : [o] --> o sqr(0) => 0 sqr(s(X)) => !plus(sqr(X), s(double(X))) double(0) => 0 double(s(X)) => s(s(double(X))) !plus(X, 0) => X !plus(X, s(Y)) => s(!plus(X, Y)) sqr(s(X)) => s(!plus(sqr(X), double(X))) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): sqr(0) >? 0 sqr(s(X)) >? !plus(sqr(X), s(double(X))) double(0) >? 0 double(s(X)) >? s(s(double(X))) !plus(X, 0) >? X !plus(X, s(Y)) >? s(!plus(X, Y)) sqr(s(X)) >? s(!plus(sqr(X), double(X))) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {!plus, double, s, sqr}, and the following precedence: double = sqr > !plus > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: sqr(_|_) >= _|_ sqr(s(X)) > !plus(sqr(X), s(double(X))) double(_|_) >= _|_ double(s(X)) > s(s(double(X))) !plus(X, _|_) > X !plus(X, s(Y)) >= s(!plus(X, Y)) sqr(s(X)) >= s(!plus(sqr(X), double(X))) With these choices, we have: 1] sqr(_|_) >= _|_ by (Bot) 2] sqr(s(X)) > !plus(sqr(X), s(double(X))) because [3], by definition 3] sqr*(s(X)) >= !plus(sqr(X), s(double(X))) because sqr > !plus, [4] and [8], by (Copy) 4] sqr*(s(X)) >= sqr(X) because sqr in Mul and [5], by (Stat) 5] s(X) > X because [6], by definition 6] s*(X) >= X because [7], by (Select) 7] X >= X by (Meta) 8] sqr*(s(X)) >= s(double(X)) because sqr > s and [9], by (Copy) 9] sqr*(s(X)) >= double(X) because sqr = double, sqr in Mul and [5], by (Stat) 10] double(_|_) >= _|_ by (Bot) 11] double(s(X)) > s(s(double(X))) because [12], by definition 12] double*(s(X)) >= s(s(double(X))) because double > s and [13], by (Copy) 13] double*(s(X)) >= s(double(X)) because double > s and [14], by (Copy) 14] double*(s(X)) >= double(X) because double in Mul and [5], by (Stat) 15] !plus(X, _|_) > X because [16], by definition 16] !plus*(X, _|_) >= X because [7], by (Select) 17] !plus(X, s(Y)) >= s(!plus(X, Y)) because [18], by (Star) 18] !plus*(X, s(Y)) >= s(!plus(X, Y)) because !plus > s and [19], by (Copy) 19] !plus*(X, s(Y)) >= !plus(X, Y) because !plus in Mul, [20] and [21], by (Stat) 20] X >= X by (Meta) 21] s(Y) > Y because [22], by definition 22] s*(Y) >= Y because [23], by (Select) 23] Y >= Y by (Meta) 24] sqr(s(X)) >= s(!plus(sqr(X), double(X))) because [25], by (Star) 25] sqr*(s(X)) >= s(!plus(sqr(X), double(X))) because sqr > s and [26], by (Copy) 26] sqr*(s(X)) >= !plus(sqr(X), double(X)) because sqr > !plus, [4] and [9], by (Copy) We can thus remove the following rules: sqr(s(X)) => !plus(sqr(X), s(double(X))) double(s(X)) => s(s(double(X))) !plus(X, 0) => X We use rule removal, following [Kop12, Theorem 2.23]. popout

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